Let $f(x) = -9x^{2}-x+8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Explanation: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-9x^{2}-x+8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -9, b = -1, c = 8$ $ x = \dfrac{+ 1 \pm \sqrt{(-1)^{2} - 4 \cdot -9 \cdot 8}}{2 \cdot -9}$ $ x = \dfrac{1 \pm \sqrt{289}}{-18}$ $ x = \dfrac{1 \pm 17}{-18}$ $x =-1,\frac{8}{9}$